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Since shock waves have non-negligible viscosity, why do the normal shock relations ignore friction/viscosity?
The normal shock relations (seen below) are derived by using steady, 1D, neglect potential, no shaft work, adiabatic and zero viscosity assumption. However, since shock waves are "thin regions of high velocity and temperature gradients, where, friction and thermal conduction play an important role," (Anderson, Fundamentals of Aerodynamics) why is the assumption of zero viscosity correct?

While I know that the normal shock relations give the correct answer, I am unsure of why neglecting friction is justified and does not lead to inaccuracies.

Neglecting friction does indeed lead to inaccuracies. Those, however, are small for weak shocks, so using the equations for inviscid shocks gives a good approximation. Especially in the times before electronic computers, this was a major simplification while the error was tolerably small.
Stronger shocks should be avoided and replaced by a cascade of weaker, oblique shocks in order to minimize losses, so for practical work there is little need for equations which include those losses. Only in academic work and the design of reentry vehicles with their blunt geometries would the precise calculation of stronger shocks be of any interest.
But how large are those losses? Wikipedia has a page on normal shock tables and the last column gives the ratio of the stagnation pressures ahead and past the shock. Since both should be the same in case of a lossless shock, the real number gives a good approximation of the energy losses incurred over the shock. The more the result deviates from 1, the bigger the losses. 1% of stagnation pressure is lost at a Mach number ahead of the shock of approximately 1.225. 10% is lost at Mach 1.587. It goes down from there: 50% is lost at Mach 2.5 and 90% at Mach 4.4. All values use a ratio of specific heats of 1.4, so they are valid for a two-atomic gas (or a mixture of predominantly two-atomic gasses like air).

If neglecting friction leads to inaccuracies, and strong shocks have are especially problematic, shouldn't neglecting friction for normal shocks (which are always strong since the Mach number after the shock is subsonic), always give inaccurate results? Also, what order of magnitude is this error? Any citations on this inaccuracy?

@NickHill: Straight shocks only are strong when the Mach number is significantly above 1. For the losses, please see the Wikipedia page on shock tables. I added that information to the answer.


text:
Why is wave drag non-zero for a flow without shockwaves?
Wave drag is often defined as the component of drag that arises due to the presence of shockwaves. Shocks result in an entropy rise and a corresponding increase in drag. However, when using Supersonic Potential Flow Theory (Ackeret Theory), the flow is assumed to be fully isentropic (i.e. no shockwaves) and it can be shown (see Bertin, Houghton, Anderson for reference) that the wave drag is non-zero for such flows. So, what is the mechanism behind why the wave drag arises in such a flow without shocks?
More Detail (from Aerodynamics for Engineers by Bertin)

Weak shock can be described by isentropic assumption fairly well.

Yes, but if wave drag is caused by the increase in entropy across a shock, then how does the isentropic assumption lead to a non-zero wave drag?

Maybe isentropic does not imply that there are no shockwaves, just that the increase of entropy across them is ignored.

The drag can be computed from Navier-Stokes momentum equations as:
$$D=-\oint_{S_\infty}p\hat{i}\cdot d\vec{S} - \oint_{S_\infty}\rho (\hat{i}\cdot\vec{u})(\vec{u}\cdot d\vec{S})$$
Usually, all boundaries other than the rear boundary (Trefftz plane) would have the same flow properties as the free-stream, and they fall out of the integral. We are left with:
$$D=\int_{S_T}{\rho u(V_\infty-u)dS}$$
As you correctly pointed out, the momentum defect of a potential flow should be exactly zero, which means that drag should also be zero.
However, for oblique shock in small-disturbance potential flow, the oblique shock (or rather, each constant potential wave characteristic) extends to infinity, carrying energy with it. If you want to calculate the far-field drag, you would need to do the integral around the entire far-field. This is different from energy dissipation in viscous boundary layer or subsonic normal shock.

Picture ref. https://nikander.github.io/compflow/Anderson/Chapter4/print/

While I see that the constant potential wave characteristic carries energy to infinity I am still confused by why drag is produced. The Mach waves which are occuring are infinitesimally weak so they should have no dissipation at all. So, why is it that when I integrate (even if I include the entire far field) that I get a finite energy loss (or drag)? After all, the flow field is still consisting of isentropic waves right?

@NickHill In the potential flow, the oblique shock is just a wave. It carries energy away from it and the net effect is drag. Take induced drag for example, there is no entropy produced nor momentum defect, but the drag is still there.

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Why is the stagnation point on the upper surface before the start of circulation?
In many textbooks, such as Anderson (Fundamentals of Aerodynamics), Bertin (Aerodynamics for Engineers), and Houghton (Aerodynamics for Engineering Students), the authors present the idea that there is a starting vortex which is formed. This starting vortex and the bound vortex that form (due to Kelvin's Theorem) co-exist with lift (as Mclean puts it in Understanding Aerodynamics). However, I am unsure on why the starting vortex forms on the upper surface. Why does it not form on the lower surface and have clockwise rotation and lead to counter-clockwise bounded circulation at the airfoil?
While I know that this is not experimentally observed nor does it make intuitive sense, I would like to know why the starting vortex does not form on the bottom.
To shed more light, here are some instances of where authors state that the starting vortex starts on the upper surface, but never explain why it doesn't start on the lower one:
(taken from Bertin):
"At the instant of starting, the flow is a potential flow without circulation, and the streamlines are as shown in Figs. 6.3 a, with a stagnation point occurring on the rear upper surface"

(taken from Understanding Aerodynamics)
"It can be shown, based on starting the flow from rest, that in the absence of viscosity, the nonlifting flow pattern of Figure 7.1.3a is the one that would occur... "

So how can it be shown (mathematically) or using strong rigorous theoretical arguments that the stagnation point cannot be at the lower surface before circulation?

The illustrations under the time-invariant, inviscid, irrotational and incompressible (potential) description is a bit misleading, in my opinion. In the real world, assuming the airfoil starts from rest (no flow field) and you start accelerating it to some airspeed, viscosity should start generating vorticity as soon as airspeed becomes non-zero. Therefore, the non-lifting flow field would never exist in the first place.
The non-lifting flow field serves better as an illustration of the flow solution using uniform + doublet flows. In this case, there is a unique solution for a uniform flow past a cylinder with radius $R$:

Radial speed: $V_r=V_\infty cos\theta(1-\frac{R^2}{r^2})$
Tangential speed: $V_\theta=-V_\infty sin\theta(1+\frac{R^2}{r^2})$

Since we can conformally transform most airfoil geometries into a cylinder, flow solutions past a cylinder is a generalization to those past airfoils. The above result assumes zero angle of attack. But since cylinder is symmetrical, a non-zero AOA simply rotates the result by that AOA, and that's why you have the rear stagnation point on the upper airfoil surface.
For a lifting flow, we would need to model with uniform + doublet + vortex flows. In this case, a set of solutions can be obtained for a cylinder:

Radial speed: $V_r=V_\infty cos\theta(1-\frac{R^2}{r^2})$
Tangential speed: $V_\theta=-V_\infty sin\theta(1+\frac{R^2}{r^2})-\frac{\Gamma}{2\pi r}$

Notice that there is, at this point, no imposition on what the vortex strength ($\Gamma$) has to be. You can have an infinite number of lifting solutions with an arbitrary value of vortex strength (at the same AOA). I think that's what (b) and (c) of your second set of figures are trying to illustrate. We get around this by specifying the Kutta-Condition and impose the stagnation point at the trailing edge to get a unique solution.
Since the total flow field must have zero circulation, the starting vortex must have a counter-clockwise circulation.

+1 and it aligns with the observed flowfield for an impulsively started airfoil

@AEhere that is an amazing photo.  I had imagined the downward deflection of the airflow wing bottom (when it was pitched up) contributes to lower pressure behind the wing, but had no idea it would form a second vortex, helping draw air over the top of the wing.  Do you think the bottom vortex contributes to an explanation of ground effect?

@RobertDiGiovanni the image I linked is of a very contrived situation showcasing a transient aerodynamic effect. It has little to nothing to do with ground effect, and everything to do with circulation.

@Jimmy You state in your reply that "...Therefore, the non-lifting flow field would never exist in the first place." But, if there is always a "lifting" flow field, how will the starting vortex ever have been formed? And moreover, under what mechanism does the starting vortex ever form?

@NickHill The circulation is started by viscosity (i.e. the airfoil boundary wall), and in large part due to the no-slip condition at the wall. You can take a look at this answer and see if it helps your understanding: https://physics.stackexchange.com/questions/135707/what-is-a-physically-accurate-explanation-for-the-kutta-condition

There are a few papers that show either experimentally or through simulation something very close to potential flow without circulation a fraction of a second after start of motion.
Before circulation forms flow off the bottom flows around the trailing edge at very high speeds (lowest pressure) and in the wrong direction forward into rear stagnation point (Highest pressure) that is in the wrong location. This creates the largest adverse pressure gradient between trailing edge and rear stagnation point on top surface.
If you draw flow separation in BL of forward flow on top surface it creates CCW starting vortex.

Nick presents a very interesting way of looking at changes in airfoil circulation, not by changing AOA, but by changing speed.  In a review of polar diagrams, it is becoming apparent that lift creation from circulation (the "Bernoulli bump" in the lift vs airfoil AOA curve), where much lift can be had with very little increase in drag, requires a Reynolds number of sufficient value.  For the same chord (of the same airfoil), this means sufficient speed to bring the stagnation point off the top of the wing to behind the wing in order to accelerate air over the top of the wing.
This is difficult to present in diagrams without picturing the motion.  As a wing, pitched to a positive AOA, speeds up, what happens is analogous to watching the transom of a boat as it accelerates to the point where the boiling wake separates from the boat and forms the "vee" behind the boat .
With a wing, once the "wake" is behind the wing, air is pulled across the top of the wing (and from all directions) to fill in the lower pressure area.  But as the wing is continuously moving forward, this is a constantly repeating process as long as AOA and velocity are maintained.
Slow the boat down, and the boiling wake returns to the transom.  It is in this light that both sets of diagrams appear to be non-contradictory.
Since the wing is pitched up, we would expect the stagnation point to form on the upper surface first.

@Nick Hill it might be easiest to model a fully symmetrical airfoil.  I suspect the stagnation point would be always behind the TE unless it was pitched.  Then the progression of it's movement with increasing speed could be observed (in a wind tunnel)

The rear stagnation point of a fully attached airfoil is always at the trailing edge.

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Does an airfoil in supersonic flow have zero circulation?
In incompressible flow, the Kutta-Joukowski theorem can be used to relate the lift (per unit span) to the circulation around an airfoil. However, if we look at an airfoil or the flat plate (see below) in supersonic flow with shocks, how can we show that there is circulation? And, if there is circulation, then how do we calculate it?

Why should it not be finite? Upper surface pressure cannot drop below vacuum, lower surface pressure cannot climb above stagnation pressure. The area is also limited. Only when you move to infinite speeds will there be a chance for infinite lift.

I should clarify: I meant to ask: Is the circulation 0 or a finite value?

Oh, the other infinite. Yes, supersonic flow can be simulated with potential flow, only what is an elliptic equation in subsonic flow becomes a hyperbolic equation so perturbations only extend in flow direction in the supersonic case. Algorithms using vortex elements exist, but for supersonic potential flow the use of doublets brings better stability, so modern potential codes don't include vortex elements. Oh, and circulation is indeed finite and not zero.

So, you mentioned "circulation is indeed finite." How can you show this?

One thing puzzles me. Supersonic flow can have no vertical component ahead of the leading-edge shock front. Clearly, air is deflected downwards in the wake. So where does the upward component of circulation come from?

Maybe I'm missing something obvious to you, but why do you think there's zero circulation?
Take the simplest case of an infinitely thin 2D flat plate in linearized supersonic flow. The solution in the perturbed velocity potential ($\hat{\phi}$) is:
$$\hat{\phi}=
    \begin{cases}
    0 & x\pm\beta z<0 \\
    \frac{1}{\beta}\alpha(x-\beta z) & 0\le x-\beta z\le c, & z>0 \text{ (upper surface)}\\
    -\frac{1}{\beta}\alpha(x+\beta z) & 0\le x+\beta z\le c, & z<0 \text{ (lower surface)}\\
    \pm\frac{1}{\beta}(\alpha c) & x\pm\beta z>0
    \end{cases}
$$
where $c$ is the chord length of the flat plate, $x$ and $z$ are the horizontal and vertical coordinates, $\alpha$ is the angle of attack and $\beta=\sqrt{M_\infty^2-1}$ is the supersonic Prandtl-Glauert factor.
The velocity field ($\vec{V}$) can  be obtained as:
$$\vec{V} = \begin{bmatrix}V_\infty \\ 0\end{bmatrix}+V_\infty\nabla{\hat{\phi}} =
\begin{bmatrix}V_\infty\pm\frac{1}{\beta}\alpha V_\infty \\ -\alpha V_\infty\end{bmatrix}$$
where $V_\infty$ is the free-stream airspeed. The first equation is plus for upper surface and minus for lower surface.
Take an infinitely thin closed path ($C$) that wrap around the upper and lower surfaces, and calculate the circulation ($\Gamma$):
$$\Gamma = \oint_C\vec{V} \cdot \hat{n} ds=\frac{2\alpha c V_\infty}{\beta}$$
This is obviously non-zero. In fact, this can be readily seen by the potential jump in the wake.
And when we calculate the lift coefficient ($C_l$) of this flat airfoil using the Kutta-Joukowski Theorem ($L'=\rho_\infty V_\infty \Gamma$), we retrieve the classic supersonic lift relation:
$$C_l=\frac{L'}{1/2\rho_\infty V_\infty^2 c}=\frac{4\alpha}{\beta}$$


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How do you calculate the wing moment coefficient knowing the moment at each spanwise station?
A lot of theoretical formulas exist for finding the moment coefficient for airfoils, but I have been confused by how to get the moments for a wing. I came up with three related questions that have been confusing me a lot:

If I know the moment coefficient at each spanwise station for the wing (so the moment coefficient for each airfoil used along the span), how can I get the moment coefficient for the whole wing? Is it just a weighted integral? If so, what's the formula look like?

How does one find the moment coefficient for a straight tapered wing?

How does one find the moment coefficient for a swept wing?

You can use the Lifting Line Theory to first calculate the downwash at each spanwise station, which are then converted to the induced AOA ($\alpha_i$).
The sectional moment coefficient and lift coefficient are:
$$c_m = c_{m_0}+c_{m_\alpha}(\alpha-\alpha_i)$$
$$c_l=c_{l_0}+a(\alpha-\alpha_i)$$
To get the total moment of a straight tapered wing (about the root chord leading edge for simplicity), it's a simple integration away, noting that the sectional moment is usually about the 1/4c of the local chord:
$$M_{LE} = q_\infty \int_{-b/2}^{b/2}{\left[ c(y)c_m(y)-\left(x_{LE}(y)+\frac{1}{4}c(y)\right)c_l(y) \right]c(y)dy}$$
where $c$ is the local chord length, $x_{LE}$ is the distance from the root LE to the local leading edge, and $q_\infty$ is the free-stream dynamic pressure. You'll have to normalize against a reference chord (e.g. MAC) to get the moment coefficient.
Finite large sweep wings have crossflow gradient that would render Lifting-Line Theory unreliable. You'd have to use a vortex surface method, such as Vortex Lattice, to derive accurate predictions.

For the total moment of a tapered wing, you calculated M_LE about the root chord leading edge. Does the moment coefficient change if you were to use any other station (say b/4 leading edge or tip leading edge)?

@NickHill Of course it will change. That's why it's very important to note where the moments are calculated.

I just noticed that your equation for M_LE has units of Moment*length. This is because the integral has c(y) and you integrate from -b/2 to b/2 and also multiply by Sref. Is the formula correct? It would help if you showed or linked the derivation.

@NickHill Thanks for spotting the mistake. There was an extra S_ref. This is pretty elementary moment summation.

I think the formula still may have a units issue. The units you have on the right hand side are Force (not moment). Right?

@NickHill Yes you're right. 2D c_m is non-dim wrt c^2, while c_l is wrt c. And in the dimensional form, their units are [Nm/m] and [N/m], respectively. So there was a missing c(y).


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Can aviation HF radios receive while transmitting?
I was listening to Gander Radio on HF at 5616 kHz today and I've noticed that operator at Gander would often start transmitting reply before the airplane finished transmission, as if trying to interrupt radio operator on the plane.
To me, this was very unusual behavior. From what I've heard and from what I concluded after listening to frequencies, it seems that HF radios in aeronautical mobile service work in traditional simplex mode, so crew normally crew wouldn't be able to hear the part of transmission send before they stop keying their radio.
So this brings me to these questions:
Are aviation HF radios capable of receiving while transmitting? If so, how?
Or if not, what's the point of Gander stepping on the plane's transmission?

Inept operator, perhaps?

I guess Hanlon's Razor applies here.

Could that be the effect of the different propagation times to your location? HF travels by reflexions between the ground and the ionosphere on daytime. I wonder if you didn't hear an apparent overlap that didn't exist.

@mins I myself do not believe that that was the case. Usually, when multiple reflections of similar strength are received, on the receiver, the sound that's made sounds a bit like an echo. I didn't hear any echos, so from this I can conclude that my reception was coming from only one major reflected component of the signal. So if the cause was delay, it would be more-or-less constant within the small time-frame of several relations.  So if I was having a delay, it would affect other relations as well, but that didn't happen.

If they are operating in simplex mode-No. However 5616 KHz  is also used by not Gander but New York, Santa Maria, Shanwick, etc. It is possible that you were listening to another ATC.

@resident_heretic Well the thing is, I regularly hear Shanwick, but in this case, I was actually listening for some time and am quite sure I heard Gander indetification numerous times. I also remember that it was interesitng for me, since I can't hear Gander regularly. The only possibility I see is that, when I was making note of the frequency, I wrote down Shanwick's frequency by mistake. It's been almost a year since I've listened, so I do not remember exactly when or how I made the note of the frequency.

@AndrejaKo:  There could be a way to hear multiple transmissions on the same frequency., Using directional antennas to transmit to different aircraft coming from different directions. The radiation pattern would would determine who hears what.

@resident_heretic Yeah, of course there are directional antennas and later or the whole beamforming story etc. But in this case, we've had communication between one ground operator and one airplane. Also, directional antenns don't seem to be very spatially selective on HF. I can quite often hear in, north Europe, stations beaming to Africa or Middle East.

No. In my experience often they can't even receive while not transmitting.

In general, the HF radios in aviation work in simplex mode.
According to the Annex 10 to the Convention on International Civil Aviation- Aeronautical Telecommunications Volume V Aeronautical Radio Frequency Spectrum Utilization Chapter 3,

3.1.1 In the aeronautical mobile service, single channel simplex shall be used in radiotelephone communications utilizing radio frequencies below 30 MHz in the bands allocated exclusively to the aeronautical mobile (R) service.

Though some aviation radios are capable of semi-duplex operation, virtually all aviation services operate on simplex.
I have no idea why the operator was stepping on the aircraft's transmission though.

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Are commercial pilots incentivized to reduce fuel consumption?
Commercial airlines spend a lot of their operating income on fuel, so in order to maximize shareholder value presumably like to use as little fuel as is safely possible.
Are commercial pilots ever given incentives (e.g. financial, performance review) for reducing fuel usage, or given dis-incentives for excessive fuel usage?

Note that flying for minimum fuel usage isn't usually the most economical way to fly (even less so as fuel prices have come down), so your basic premise is somewhat flawed.

See this answer of mine for things to consider: http://aviation.stackexchange.com/a/1860/69

There have been, at various times, and with various carriers, incentive programs set up to encourage pilots to save fuel, improve on-time performance, or some combination of the two.
An example of a fuel-saving incentives program existed at the former Continental Airlines in the early 90s, where pilots would be paid a bonus for using less fuel than planned. The problem was that it wreaked havoc with the schedule itself, as pilots would simply dial back to economy cruise thrust settings, saving fuel but taking noticeably longer to reach their destinations.
An example of an actual-vs-scheduled block incentive program intended to help with on-time performance existed at Silver Airways in their Beech 1900D fleet, whereby pilots would be paid a bonus for each minute they were able to shave off a baseline target time. Ironically enough, it was called the "Green Incentive Program" (don't ask). And it was a can of worms.
I vaguely remember other airlines experimenting with incentives programs relating to fuel consumption and on-time performance, but it's been several years since I left the industry and can't recall any other examples.
In short, yes they do get implemented fairly regularly, but they can be very problematic due to the difficulty in properly designing and managing them (incentivization is a complex problem).
Lastly, since another poster has brought this up, dispatchers do indeed perform most, if not all the flight planning, but the PIC (i.e. Captain of the flight) has the option of overriding most everything the dispatcher plans and arranges for (subject to the carrier's Operational Specifications), including amount of fuel to be carried. Basically, if the Captain decides that the plane is to be filled with as much fuel as it can carry up to MTOW, that's what's going to happen. Might not be the most business-savvy (or even prudent) decision, but it's part of their prerogative as PIC (i.e. the person ultimately responsible for the craft, crew, passengers, and goods onboard).

Seems a combination of the two types of programs (Save Fuel, and Save Time) would be more ideal. Since fuel presumably costs the company more than a few minutes late (or on time) arrival, it could be weighted to favor fuel savings, but penalize (reduce bonus pay-out) when planes arrive late. This would give pilots the incentive to find fuel saving opportunities, but also not delay arrival too much.

Saving "time" can present problems all of its own, e.g. get to your destination too early and have ATC put you into a hold at (gasp!) low altitude until your landing slot opens up. Ideally you'd want to set up a system to encourage saving fuel while keeping as closely as possible to the planned schedule. From what I've seen the best way to achieve that has less to do with monetary incentives and more to do with optimal flight profile planning, and pilot training, education, advocacy for fuel-efficient operating procedures.

The problem with being "a few minutes" late on each leg is that it adds up throughout the day. You might be, say, 10 minutes late on the first leg of the day, but, if you keep doing that, by the end of the day you'll be running an hour or more behind. At that point passengers have missed connections and have to be rebooked (or put up overnight), your crews might have exceeded duty day limits, you might even have had to pull in reserve aircraft and crews to try and get scheduled flights out on time. The costs (both direct and lost revenue) of chronically delayed operations are no joke.

@SnakeDoc, the biggest problem is that incentive for either puts that above safety, for which it is almost impossible to come up with reasonable incentive. If you give bonus for being on time, pilots will be more willing to try to salvage poor approaches where go-around would be prudent, because go-around is 10–15 minutes delay. If you give bonus for saving fuel, pilots will be reluctant to take much contingency fuel and then plant themselves in corners when delays happen. But if you try to fix it by giving bonus for not having incidents, pilots will just not report them, which is even worse…

It's seems like you are talking about part 121 operations which is like delta to Chicago and Atlanta and daily passenger operations. The pilots don't really do the weight and balance(fuel required). Dispatchers do the calculations and then it's sent to the pilots and they double check the math.
I'm gonna say they don't because it's already at mins.
Then you have other factors like it's cheaper to buy fuel at this airport so I will top of the tanks so you can fly to another airport and back.

But carrying excess fuel like that makes the plane heavier, which increases fuel consumption. Is this really worthwhile for any but the shortest hops?

@TonyK turnaround times are also an issue when it comes to fuelling up.

@TonyK Yes, they do that if the price difference is enough to justify it. It's called tankering. I'm not sure when it's advantageous but I suppose fuel prices can be radically different between different countries for political reasons.

Fuel prices can be different even within the same country, especially in a place like the US where different states can have different levels of sales tax and/or excise duties, and that's before considering situations where an  operator could negotiate discounts with a single fuel provider for bulk purchases at an individual location.

After the collapse of the USSR and break-up of Aeroflot into smaller regional entities, there was (reportedly) a widespread practice of such incentives. Or rather, disincentives to 'waste' fuel. Say, a pilot could be penalised for doing such 'unnecessary' thing as a go-around. This was often unofficial or semi-official, but still put pressure on pilots to compromise safety. It is believed such practice may have contributed to some accidents in the 90s.

Incentivising pilots for reducing fuel requests was definitely not the done thing in Australia when I worked at a commercial airline. The responsibility and final decision making of the captain was absolute. For those pilots perceived to load up too much 'granny gas' as it was called, we used to make charts showing them where they were ranked amongst their fleet type colleagues for putting in more fuel than planned. For the 'bottom' say 5-10% of 'overloaders', they were called in for a meeting with the fleet captain, and shown where they sat in the ranking tables. The pilots found the process endlessly fascinating, but both the company and the crew were aware these conversations were not 'directive' in anyway, just a 'hey here's where you sit: do you think our planning parameters are perhaps too tight?' or 'have you just been getting many marginal weather flight rosters lately?'. This was almost always enough to get the outliers to change their ways slightly (we tracked that too). But at no point were they incentivised to reduce their additional fuel requests, and no-one in management would even ask: the tech crew's charge over their aircraft on the day of operation was absolute, and this was understood by both management and the pilots. Everyone felt that this was an important part of a good safety culture, and never under question, even during tough financial times. Had any bozo tried to tell a pilot how to plan their gas, they'd have been given very short shrift by the pilot concerned, but I'm not aware of that even happening, despite the presence of some genuine 'overloaders'!

I did however find this article recently: an allegation that the pilot cancellled another go-around due to company fuel-saving opportunities: https://www.abc.net.au/news/2007-04-12/garuda-crash-pilot-may-have-been-trying-to-save/2239656

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text:
How is horsepower measured for aviation piston engines?
As far as I can tell, aviation piston engine power output is typically measured in HP (horsepower).
Automotive engine power output is typically measured by running the host vehicle on a rolling road, or dyno or dynamometer.  This directly produces a torque curve, which may be integrated with respect to RPM to provide a power curve.
Obviously it wouldn't make much sense to put a piston-engined aircraft on such a device.  Instead, how is horsepower measured for aviation piston engines?

You can put engines on a dyno hooked up to the shaft directly.

Like anything else a Dynamometer is used except the engine is bolted directly to it (most likely on a test bench). In a car a rolling road is simply a way to couple the tires to the dyno without having to have any custom fittings. For what its worth cars lose a bit of horse power in the transmission/drive shaft/differential so shaft horse power and wheel horsepower are not always the same. However in a plane the prop is bolted more or less to the end of the crank (with the exception of some planes with stepper gear boxes). In any case shaft horsepower on a plane engine is more or less whats available at the prop.
Here is a lycoming engine on a dyno

(source)

Also, long before the first engine is installed in a car on the assembly line, various pre-production examples are run on a dyno in exactly the same fashion.

How is horsepower measured for aviation piston engines?
One idea is to use a "shaft power meter".

Measuring it is one thing, but you actually have to absorb the power output somehow. And spinning a big propeller in a small room is not the best way of doing this.


text:
About ramjet efficiency in supersonic speed. Is there an optimal speed?
I know there is an optimal speed for ramjet.
This are the results of my research.
The reason the efficiency increases with supersonic speed to an optimal value then decreases is related to the combustion speed.
The burn flame in the combustion chamber propagates at subsonic speed.
When the air is compressed the speed of intake airflow falls subsonic while flowing into combustion chamber. However, when the air speed in the combustion chamber matches the flame advance speed, then the flame remains stable inside the combustion chamber and the top efficiency is achieved.
At higher speeds the airflow still needs adjusted to match the flame propagation speed. This means the compression has to reduce the speed to the airflow to the same subsonic value, while dealing with even stronger effect generated by shockwaves. There is a maximum flow rate through engine that still maintains the maximum subsonic speed of the flame propagation. Anymore air resulted from higher speed will push the front of the flame into the exhaust nozzle which in turn pushes the expanding exhaust gas outside of the nozzle, with loss of thrust.

Well, that would depend on the design parameters, but for example this ideal ramjet would, with stoichiometric combustion process,  provide best thrust per unit mass flow at a speed of approx. 2.4 Mach

...the best operating range of a hydrocarbon-fueled ramjet is $ 2 \leq M_0 \leq 4$ . The parameters used are $ \tau_\textrm{max}
= 10$ , $ a_0 \approx 300\textrm{ m/s}$ in the stratosphere, $ f_\textrm{stoich} = 0.067$ for hydrocarbons, such that $ a_0/g
f_\textrm{stoich}
\approx 450\textrm{ s}$ .

Thank you. It answers the question


text:
Can a VTOL jet such as the Harrier enter a Vortex Ring State?
I'm curious, because they obviously don't have those huge rotors pushing the air away, but they are still getting their upward thrust from somewhere.
My gut says, yes, but it is more difficult. Is this right?

I don't believe the Harrier can since it uses its main engine and runs the thrust through ducts to produce lift. There is no rotor to stall as you descend through your downwash. The V-22 or F-35 may be a different story because it does use a lift fan or tilt-rotor.

The main problem with VRT is the airflow through the rotor which is disturbed because it is ingesting its own downwash.

The large area of air accelerated downwards keeps being sucked back in to the rotor because the helicopter sinks back into it. In this state the rotor provides very little thrust - it acts as a sort of divide-by-zero state, as the transition between normal flight and windmilling when the air streams upwards.
The flow field of the Harrier is different:

Thrust is produced inside the engine, which is shielded from the effects of aircraft vertical velocity: air is sucked in through the inlet next to the cockpit.
The air stream velocities are higher. Air is accelerated downward in a much smaller area at much higher velocities. There is no fuselage which partially decelerates the airstream, like in a helicopter: the downward flow field is much cleaner.

The defining problem of VRS is the lack of thrust from the rotor at downward velocities, this would not occur with the shrouded jet engine of the Harrier.

I would imagine the VTOL version of the F35 would be more at risk due of vortex rings due to the fan

"Does never seem to be"... seems to be missing something.

No.  Considering that lift is created by the downward wash of jet exhaust as opposed to air flow through a powered lift disc, this phenomena isn't encountered here.
Harriers do have a problem called blowback where hot section exhaust can be ingested into the engine, causing a flameout in cases of high descent rates or where the landing pad becomes flooded with exhaust gases, causing them to be reingested into the engine.


text:
How useful would the space shuttle OMS actually be in atmosphere?
When speaking of space shuttle landings, they always mention that it is "very poor glider, with one shot, no go-arounds."
Of course this is true; however, if the Shuttle found itself in a situation where it was a little short on delta-v to reach land/a runway (perhaps during an RTLS abort), would the OMS make any meaningful difference in extending gliding distance?

Not all that useful - the OMS engines had 6,000 lbf of thrust each.  For comparison, the Allison J35 (used in the F-84) produced 5,600 lbf of thrust.
Comparing weights, the shuttle weighted 165,000 pounds empty (reference) while the F-84G weighed 11,470 pounds empty.  Since we have two OMS engines on the shuttle, we're double that to 22,940 pounds. So we have to push approximately 8 times the weight.
There are a few other considerations - The OMS engine only had enough fuel for 1,000 ft/sec of delta-V, fuel would be needed for entering orbit, and later 200 ft/sec for reentry, along with whatever else was used for maneuvering during the mission. Therefore there probably wasn't a whole lot of fuel remaining. Also in my comparison don't forget that the shuttle is not known for being very aerodynamic, so you might be able to manage small changes, but nothing of significance.

Actually it's even worse. The thrust and delta V are for vacuum operations, at 1ATM it would be significantly smaller

A vacuum engine at sea level won't just suffer loss of thrust/Isp, it might be damaged by the shock waves inside the nozzle bell.  But I hadn't realized there was so little dV available from OMS -- I'd thought it was more like twice that figure.  And 1/14 G in vacuum -- hardly enough to spill the coffee.


text:
What is the simplest way to explain a sonic boom that is still technically correct?
I'm looking for an explanation that preserves technical accuracy while remaining brief. Bonus points for including the double boom.
Try to avoid discussing laminar flow and such.

What is not covered in the Wikipedia article?  Asking for a technical correct answer without being technical is surely a tautology?

By still technically correct, I mean doesn't tell any physics lies or lying by omission.

And our mission is to build our own database of answers to aviation questions, not google search skills.

I fully understand that, but I have no idea how to answer your question.  An analogy would be "please explain electricity without mentioning electrons or coulombs.  I have even less idea of what you are asking for following your comment.  What do you mean by "physics lies"?  Are you disputing the accepted explanations for what causes sonic booms?

I can explain electricity without mentioning that. Consider electricity like a fluid in a pipe. Voltage is like the water pressure. Current is like the amount of water flowing. A battery is like a pump, and a faucet is like a bulb. And no, I am not disputing accepted explanations for sonic booms, I am just trying to avoid answers like "airplanes fly because the air has to move faster over the top of the wing to get to the back at the same time", which is patently false.

Indeed, but that is not the accepted explanation for how lift is generated.  Anyway, I've tried below.  Is this what you're looking for?  I still don't understand what you are asking for.  Your explanation of electricity is most unsatisfactory.  It does nothing to explain why water is not like electricity.

What If I compared the sound waves to bow wake from a boat?

@Simon And I am aware that my water analogy is weak; it was not my intention to actually give a comprehensive explanation, just an example of the path I would take.

I will try to explain the principle, briefly and practically as you requested.
Imagine that we have an airplane flying in position 1. It is flying at supersonic speed.

Good, now this airplane moves forward. But it is not the only one moving forward, the perturbation that the airplane produces over the air moves as well, but at a different speed, the speed of sound. In the next figure you will see that the perturbation (or pressure wave) moves less distance than the airplane, which moves from 1 to 2.

The airplane keeps moving to position 3, and at position 2 there are some perturbations (pressure waves) that are also generated and also move as speed of sound. However, I have smartly selected a perturbation that reaches exactly the same point in space that the perturbation (pressure wave) that I selected is reaching as well. So both perturbations meet at the same point.

As you imagine these two are not the only perturbations (pressure waves) that are meeting at this point, there are several points between 1 and 3 that also generate perturbations that join. Actually, this is also not the only point in space where perturbations (pressure waves) join together: there is a line that envelopes the points where the pressure waves meet at each instant. Well, actually a surface in 3D.

OK, if you do the same with a better painting tool than me… you draw the same but also evolving with time like this:

Which is the image in Wikipedia. What is the effect that you experience?

So the orange observer that is located in advance/before the blue lines does not feel any perturbation due to the airplane, nor hear it, as the pressure wave have not reach him yet.
However, the green observer has already felt the airplane passing by and sees how the air is altered by the airplane (imagine that you are close enough).
Finally the brown observer is just in the middle, coming from a region which has not yet experienced the perturbations made by the airplane and finds suddenly a concentration of perturbations (pressure waves) as described above. That is a sudden change in air flow properties and is experienced as a sonic boom.
To give a real end to this, notice that the further you are from the airplane the less strength the pressure waves have when they arrive at the blue line, making the sonic boom less and less intense.

Is this what you're looking for?

When an object passes through the air, it creates a series of pressure
  waves in front of it and behind it, similar to the bow and stern waves
  created by a boat. These waves travel at the speed of sound, and as
  the speed of the object increases, the waves are forced together, or
  compressed, because they cannot "get out of the way" of each other,
  eventually merging into a single shock wave at the speed of sound. In
  smooth flight, the shock wave starts at the nose of the aircraft and
  ends at the tail.
There is a rise in pressure at the nose, decreasing steadily to a
  negative pressure at the tail, followed by a sudden return to normal
  pressure after the object passes. This "overpressure profile" is known
  as an N-wave because of its shape. The "boom" is experienced when
  there is a sudden change in pressure, so the N-wave causes two booms,
  one when the initial pressure rise from the nose hits, and another
  when the tail passes and the pressure suddenly returns to normal. This
  leads to a distinctive "double boom" from supersonic aircraft.


text:
Why are there no supersonic propellor driven aircraft?
Pretty self explained in the title.
Yes, I realize there is no widespread demand for them; this is not what I'm looking for in an answer.
I'm looking for an answer that covers more why they can't be built (or aren't practical to). I know that jet engines may generally be better, but I've recently been doing reading on aircraft like the Thunderscreech and it's made me curious why exactly it won't work.

The blade has 0 velocity  near the center and maximum velocity at the end, so somewhere in the middle you will get hit by the bad aerodynamics of transonic. Maybe you can develop a ring shaped propeller that operates mostly in the supersonic region?

The NASA has done research on the supersonic propeller at the end of WWII, which can be found here and here.
The main reasons that we don't see any supersonic propellers are the following:

The supersonic regions at the blades create tremendous forces which will tear the propeller apart. Thus the design needs to be such that these forces can be mitigated. This is also the case for jet engines, which is why the ducts are designed such that the airflow into the engine is slowed down to subsonic speeds.
The developments in jet engines were going at such a fast rate that the propeller was soon surpassed and was written off. However, since turboprops are much more efficient than jet engines they might pop up on the research radar in the near future.


text:
Why Clyde P. / Boeing B-52H Stratofortress flight from 1962 took the longer route?
Third on list of longest non-stop, non-commercial powered aircraft's flights is January 10–11, 1962 flight on Boeing B-52H Stratofortress made by major Clyde P. Evely and crew. Can anyone provide any details about this flight?
Wikipedia says, that it was a flight from Japan (Okinawa) to Spain, via Tokyo, Seattle, Fort Worth, Washington and Azores. I'm especially interested in knowing, why that way? Most (all) of today commercial flights are flying Europe to Japan / Far East the opposite way, that is through Asia.
Is there any source of information, why did that flight took the longer route? Was that specially made to break the world record or was there any other circumstances to take such decision?

Would that shorter route have been overflying the USSR and/or the PRC?  In 1962, a B-52 doing that might have been poorly received...

Well... that could be a good point!

relevant: http://aviation.stackexchange.com/questions/14920

The B52 was not going to go that far without mid-air refueling, that route ensured that there was mid-air refueling available, and plenty of friendly places to land in case of trouble.

@GdD Cited Wikipedia article have in first paragraph a notice saying: "This list of flight distance records contains only those set without any mid-air refueling" (emphasis mine). I don't know nothing about "friendly places to land", but in conjuction with Ralph's comment this could be the point.

I remember in 1991 taking a direct flight from London to Tokyo over the USSR.  AIUI the ability to do that was new; 12 months before a previous group on the same trip had used a longer route over the Pole.

@NigelHarper It's also a little easier to get permission to fly a passenger airliner over another nation than it is to get permission to fly a B-52 over it...

@reirab Exactly. And if it was only just possible to get permission for commercial flights to overfly the USSR in 1991 it shows just how impossible it would've been to get the same for military craft 30 years previously.

The B-52 Stratofortress is a military jet, and specifically a US military jet, which if flown over USSR was very much in risk of being harmed. Thus, it was an intelligent decision to follow a longer route, rather than being a victim of ongoing cold war, following the longer route could have also helped the crew in achieving the longest flight record. Secondly, the B-52 did not have a very good history even before 1962, thus taking precautions was a necessity for the crew, and they did this by choosing to fly over more helpful nations.  Also, aviation was not developed very much in Asia back in 1962 with very few strips of a large distance, to be used in case of emergency.
For more information, you may refer this and this!

The B-52 is a military jet. It's still very much in service today. It's one of the longest-serving military aircraft in history, especially for a jet.

Saying that the B-52 would have been at risk of being harmed if it overflew the USSR is something of an understatement. Doing that might have started World War III.

Okay , then as @RalphJ says, it would at least have been poorly received.


text:
What is the third airport near Dubai visible on Google Maps?
I know Dubai International (DXB) and Dubai World Central (DWC). What is the third Dubai Airport, not named on Google Maps?
Is it military, GA or commercial. Why there is no clear access road to it?

Zooming in a little more, it identifies the airport as Al Minhad Air Base. Turning on Google Earth view, you can see a single road leading to the airbase from the east.

The airport is Al Minhad Air Base:

Al Minhad Air Base is a military installation in the United Arab Emirates. The base is located approximately 15 miles (24 km) south of Dubai and is operated by the United Arab Emirates Air Force. It is currently the headquarters of Joint Task Force 633 and supports Australian operations in the Middle East.

If you turn on Google Earth view, you can see the single road leading into the airbase from the east:

Camelicious :)

I was curious as to what it was but was afraid to look it up.

Thank you @Steve for adding the link and making the day just a bit brighter!

@Steve Ooops, I am too late! :( "PAGE NOT FOUND" (http://camelicious.ae/home.php).

@trejder The 404 page still includes the text "Over 35 years ago, the idea of a camel-milking facility was born at the Central Veterinary Research Laboratory (CVRL) in Dubai", "Emirates Industry for Camel Milk & Products" and the "Emirates Camel Goodness" logo. So it appears to be a camel dairy, for want of a better phrase.


text:
Has any airliner ever collided with a glider?
I've read about many near-collisions between gliders and airliners, but did such a collision ever happen in mid-air, actually?

I would imagine it happened a considerable number of times during WWII given that the skies were packed with towed gliders.

Packed with towed gliders, yes, but not also with what we'd call airliners.

Yes.  (I found this by Aaron Holmes's hint that he searched only English sources.)
On 1999 Feb 12 at 3:40 pm, a Grob G103 sailplane F-CGXB (possibly "in wave flight") collided with an Airbus A 320 F-GJVG at 8600 feet, near Montpellier, France.  Both landed safely with minor damage.  There were no casualties.
Summary: https://aviation-safety.net/wikibase/24140
Accident report, excerpted below:  http://www.bea.aero/docspa/1999/f-xb990212/pdf/f-xb990212.pdf

L'Airbus A 320 à destination de l'aérodrome de Montpellier Méditerranée est
autorisé à descendre vers le niveau de vol 80. Dans le même temps, en vol local
autour de l'aérodrome de Saint Martin de Londres, le planeur G 103 évolue vers
8 600 pieds dans une ascendance au-dessus de la montagne de la Seranne.
Pendant la descente, l'équipage de l'A 320 aperçoit le G 103 et effectue une
manœuvre d'évitement par la droite. Le bord d'attaque de l'aile gauche de l'A 320
et l'extrémité supérieure gauche de l'empennage du G 103 se heurtent. Les deux
appareils se posent sur leur aérodrome de destination.

L'extrémité arrière gauche de l'empennage du G 103 est arrachée.
Le bord d'attaque du bec n°4 de l'aile gauche de l'A 320 est enfoncé et déchiré.

English translation:

The Airbus A320, with destination of Montpellier-Méditerranée Airport, was authorised to descend to flight level 80.  At the same time, in local flight around Saint-Martin-de-Londres airport, the G103 glider rose to 8600 feet in a climb above the Séranne massif.  During the A320's descent, its crew spotted the G103 and executed an evasive maneuver to the right.  The leading edge of the left wing of the A320 and the upper left extremity of the empennage of the G103 collided.  Both craft landed at their destinations.

The rear left end of the G103's empennage was torn off.
The leading edge of slat #4 of the A320's left wing was dented and torn.

Bien joue', M. Goudeseune. Very well discovered!

I have attempted an English translation of the excerpt you quoted (edit approval pending.)  Feel free to modify it or roll it back if you think I've totally botched it.

wave flight HAZARDS!  whoa!  https://www.williamssoaring.com/pilot-info-wave.html   they should add "smack in to an Airbus"

@Fattie, well, most of that document before the hazards section deals with mitigating that very hazard, that is getting clearance into the class A airspace where large jets normally fly.

Just wow. How must it feel to touch a flying Airbus? And the turbulence behind it? Bigger planes have succumbed to that.

That's crazy that the glider survived!

On 28 August 2006, a collision occurred between a business jet and a glider near Reno, Nevada, at approximately 16 000 feet. See here for details.
To partially quote from the aviation-safety.net summary at the link above:

The Hawker 800XP departed Carlsbad (CLD) on a flight to Reno (RNO). Descending for Reno, the flight crew was cleared by air traffic control from 16,000 feet to 11,000 feet. Just prior to commencing the descent [...] [T]he Hawker jet impacted the right wing of the glider near the outboard wing joint, shearing off part of the wing. The glider entered a flat spin and the pilot bailed out. [...]
In the collision, the nose section of the Hawker received substantial damage [...] The crew [continued to ..] Carson City, and elected to land there. As the flight neared Carson City the flight crew noted that the right engine shut down as a result of the impact.

The NTSB published accident report number LAX06FA277A. The business aircraft registration was N879QS.

SE discourages link-only answers in case the link stops working (as it has in this case).


text:
Velocity induced by a straight vortex filament
Consider a straight vortex filament as shown below. At each point, there's a point vortex of strength $\it T$. Consider that a point $P$ is there on the outermost circle of flow induced by the vortex about the origin $O$. That is the point $P$ lies in the plane containing the flow induced by the vortex at $O$. So, the only point on the vortex filament that will induce a flow at point $P$ is $O$.
Therefore, the vortex at point $O^{'}$ will not induce any flow at point $P$. Vortex at point $O^{'}$ can only induce flow at those points that are in the plane containing the flow induced by the vortex at $O^{'}$.

So, my question is how Biot-Savart Law is applied for a straight filament vortex? How can the segment $dl$ induce a flow at a point $P$ that is not in the plane? Only the vortex centered at the point of intersection of line and perpendicular from point $P$ to the line can induce the flow at $P$. Please clear my query.

You need to think in 3D. A vortex will induce a flow speed everywhere.

That is the point  lies in the plane containing the flow induced by the vortex at

Why would you think that vorticity induces a speed only in one plane? You start from the wrong presupposition.
Now consider the rings around the vortex filament in your first drawing. All around those rings a flow speed is induced. Its magnitude drops with the inverse of the radius of one particular ring but is constant along the circumference of that ring. Actually, that ring is only a section through an infinitesimal tube, just as infinitesimal as that vortex filament.

Irrotational vortex. © Silver Spoon, picture source.
Also, the symbol for vortex strength is a capital Greek letter Gamma ($\Gamma$), not $T$.

Thanks for the response! Now, I understood that I had a wrong idea about the 3D vortex filament.

@Peter vortex induce velocity because of viscosity? if you start rotate spoon in the middle of bucket of water, after some time all the water will spinning in that direction..so at zero viscosity induction is zero? Is this theory talk about that?

@user628075 vorticity ≠ viscosity!

@PeterKämpf So what is mechanism how vorticity induced velocity?

@user628075 That is the part which makes potential flow theory so unintuitive. The idea is that it works like electromagnetic induction, following the same laws. Of course, in reality viscosity is needed, but potential flow theory assumes inviscid flow. I only wanted to point out the mistake in your comment. The answer nowhere refers to viscosity.

@PeterKämpf How this theory works if that is not what is happening in real fluid? This is unbelievable

@user628075 Please leave questions of belief to your religious authority. Here we deal with a theory which helped to explain and predict flow effects with little computational effort. It did its job when nothing else did. And still does today.

@PeterKämpf yes it  deliver results with maybe 60% accuracy


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Turbulent flows over 6 series airfoils
6 Series Airfoils were developed to enhance the laminar flow region over the airfoil. But, the Reynolds number at which planes fly are high for the flow to be Laminar. Then what's point of introducing 6 series airfoils when laminar flow is not possible?

Flow at the forward stagnation point of an airfoil is laminar, because the local velocity will be zero. The boundary layer originating form the stagnation point will usually not remain laminar for long, but some airfoils like the NACA 6 series were optimized to conserve this laminar flow as far down the chord as possible.
A laminar boundary layer is thinner and offers less drag, so it is desirable for aerodynamic efficiency. One of the more famous laminar wing designs was the P-51 Mustang, which used the NACA 66-(1.8)12 (a=0.6) airfoil at the wing roots.
In reality the boundary layer will transition at some point before reaching the trailing edge, and become turbulent. In the case of the P-51, a shot down example was analyzed by H. Schlichting. He found that the transition point was at the 50% chord point for $Re=4\cdot10^6$ and as far forward as 20% of the chord for $Re=7.5\cdot10^6$, which suggests that the Mustangs performance had more to do with the polished finish and flush rivets of the wings than with the airfoil used.

Thanks for replying to my question. Here is my question : how local velocity being zero makes boundary layer velocity zero ?

@Pavan what non-zero velocity value would you expect? You can prove that it must be zero at the stagnation point by taking the limits from above and below the stagnation point. Otherwise I am not fully understanding your question.


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Cruise Mach Number less or more than Critical Mach Number
Do commercial airplanes cruise at Mach number higher than the critical Mach number? And what is the critical Mach number of airplanes like B777 or A330?

Given that exceeding critical mach in an airplane not designed for supersonic flight can lead to departure from controlled flight I'd say they probably cruise below critical mach.

I guess by exceeding Drag Divergence Mach Number (Mdd), there is a significant rise in the drag. There won't be any issue if the plane is flying at Mcrit or in between Mdd and Mcrit, right?

@GdD critical mach is the lowest airspeed at which the airflow reaches Mach 1 over some part of the airframe. That does not necessarily mean departure from controlled flight. In this video you can clearly see the normal shock wave dancing back and forth across the wing as an A320 exceeds critical mach in cruise flight. And on a 737. And on a 757.

@Pavan According to this document, the max cruise Mach number for most of the airliners cited is simply equal to the drag divergence Mach number, Mdd. But that is not  the same as Mcrit.

@Fiddlesticks Mdd is more than (typically more by 0.1) Mcrit. But, flying very close to Mdd will increase drag significantly. So, don't you think it is better to fly between Mdd and Mrit?

The critical Mach number $M_{crit}$ is defined as the lowest Mach number at which the airflow over some point over the airfoil reaches supersonic speed, that is, Mach 1. Usually, flying greater than $M_{crit}$ will produce a normal shockwave perpendicular to the flow, at the point of minimum pressure on the airfoil. I must note that if flying slightly above $M_{crit}$, there is a possibility of the isentropic recompression of the fluid, which means that no shockwave will be formed and thus no losses in form of the entropy will occur. This is especially the case with supercritical airfoils flying slightly above $M_{crit}$.
Critical Mach number for a typical transonic airliner is not equal to divergence Mach number $M_{dd}$. In fact, many aircraft will exceed $M_{crit}$ during the cruise, which you can see on the following video:

. I believe that the average value of $M_{crit}$ is around $0.78$ for today's airliners like A320 or B737.
Exceeding $M_{crit}$ does not mean the aircraft reached $M_{dd}$ or is about to lose control. Control difficulties may be possible if $M_{mo}$ is exceeded.
For A320 and B737, the maximum operating Mach number $M_{mo}$ is $0.82$. This is primarily due to the buffet (and ultimately high-speed stall/shock stall) and (or) flutter occurrence. Exceeding normal speed range may cause flutter to appear. It also may put the aircraft in grave danger if shock stall occurs. Shock stall will occur if the shockwave formed on the airfoil is too powerful and separates the boundary layer. In this case, the airfoil pressure distribution changes and introduces an intensive nose-down pitching moment. To avoid such scenarios, the aircraft are limited with Mach number as well. There might be some corrections on the $M_{mo}$ from the divergence Mach number standpoint, but $M_{dd}$ is not the primary reason why such aircraft are limited with respect to Mach number.

Exceeding Mmo cannot cause flutter to occur (by regulations). I think what you mean to say is buffet.

What I meant to say is exceeding the normal speed range may cause flutter to appear...
Transonic buffet will always be present if the aircraft is experiencing shock stall, since these two phenomnenos are very closely related.

Flutter is possible only if you exceed Mdf, which is an extra margin over Mmo. You won't encounter shock stall up to Mdf either, as they've been flight tested there. Buffet is the only thing you will experience above Mmo.

As you know, a transonic buffet is a condition where a separated turbulent boundary layer strikes the airframe with considerable force causing high amplitude vibrations. On a swept-wing aircraft, the separation will start at the wing root and work its way out towards the wingtips. As the first signs of separation will take place at the wing root, the transonic buffet will act as a consequence to "warn" the pilot to stay out of such high velocities.

If the speed increases further, more wing will suffer from boundary layer separation and buffet will become even more severe until reaching the point where expressed nose down tendency will occur or result in the possible structural damage of the aircraft.The point is, the buffet acts as a warning sign to stay out of a shock stall, buffet should not happen in the first place,if it happens,this means that the shock stall is already gaining strength until it becomes impossible for the pilot to control the aircraft. Staying out of shock stall means eliminating the buffet in the first place.

For a well designed wing, shock stall should not occur, at all, for the range of AOA of interest (unless through excessive maneuvering). Buffet, however, occurs with or without shock stall. Part 25 is written specifically to seek out buffet boundary, as well as deterrent buffet up to Mmo (25.251e). Level flight up to Mdf must be free of excessive buffet up to Mdf (25.251b).

AC 25-7C section on buffet testing requires having no stick force reversal up to deterrent buffet or 2.5G maneuvering. Simply put, there is ample margin in both airspeed and maneuvering prior to encountering shock stall or flutter (which is a totally different topic).

In my opinion, I think that shock stall starts when boundary layer separation occurs and decreases the coefficient of lift. Those things will occur as the transonic buffet starts, but at first it will be negligible. Only when enough boundary layer is separated, the pitching moment will change enough to be felt on the control column and etc.

Depending on where the shock separation begins, the coefficient of lift may not decrease at all through to deterrent buffet and beyond.

Why do you think that the coefficient of lift will not decrease in the buffet, could you please elaborate on that? I must point out that on a swept-wing aircraft, the boundary layer separation will start at the wing root where most of the lift is generated.

Because of past (proprietary) work in wind tunnel tests on several high speed airplanes? Because of the stick force/G flight testing up to deterrent buffet? Because of AC25-7C requirements?

I highly disagree with the statement that the lift coefficient will not decrease due to the boundary layer separation at the transonic buffet. Any book about the Transonic Aerodynamics cannot back up the given statement.

Take it up with FAA/TCCA/JAA and flight test groups. I suggest you give AC25-7 a read.


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What is a jet (unit) shown in Windows 10 calculator?
From Windows 10's calculator:

What is a "jet" (physical unit)?
And why does it equal 480 knots or 888,88 km/h, if a jetliner's cruising speed is around 950 km/s? Or why does it equal 0.73 M, if a regular jetliner's cruising speed is 0.85 M?

+1 for cursing speed stated in km/s. That's actually plausible.

I think "950 km/s" is a typo; surely the jet isn't moving at 950 kilometers each second.

Laden or unladen?

European or African?

Thank you all for all your enlightening comments. It is really a nice thing that such a silly (after all) question made that number of people laughing and happy! :> There's one thing that I always had trouble to understand. If this question is OT (while it clearly is, since it asks about non existing unit) why does it have so many upvotes and two, quite professional answers? :>

@trejder Being OT doesn't mean it isn't funny.  The OT flag is, in large part, so others don't use this as an example of topicality.  It doesn't stop people from answering if they want to despite it being OT.

@StephenS well, closed questions cannot receive new answers, so yes, it does stop people from answering. it is simply not retroactive.

It is not a unit. It is just Microsoft trying to be funny. Or to convey an idea of the magnitude.

Thanks @Jackie for pointing out that around March 2019, Microsoft has released this calculator as open source under MIT license. It is available in this Github repository. The source code sheds light on the definition of these pseudo units:
Description in ./src/Calculator/Resources/en-GB/Resources.resw:
<data name="UnitName_JumboJet" xml:space="preserve">
   <value>jumbo jets</value>
   <comment>A jumbo jet (eg. Boeing 747), used as a
            comparison measurement unit for length.
            (...)
    </comment>
 </data>
...
<data name="UnitName_Jet" xml:space="preserve">
   <value>jets</value>
   <comment>A jet plane, used as a
            comparison measurement unit for speed.
            (...)
    </comment>
 </data>

Conversion coefficients in ./src/CalcViewModel/DataLoaders/UnitConverterDataLoader.cpp
/*categoryId, UnitId, factor*/
static const vector<UnitData> unitDataList = {
        ...
        { ViewMode::Speed, UnitConverterUnits::Speed_Turtle, 8.94 },
        { ViewMode::Speed, UnitConverterUnits::Speed_Horse, 2011.5 },
        { ViewMode::Speed, UnitConverterUnits::Speed_Jet, 24585 },
        ...
        { ViewMode::Length, UnitConverterUnits::Length_Paperclip, 0.035052 },
        { ViewMode::Length, UnitConverterUnits::Length_Hand, 0.18669 },
        { ViewMode::Length, UnitConverterUnits::Length_JumboJet, 76 },
        ...
};

Apparently these definitions assume length in meter, and speed in cm/s. Thus, in Redmond the typical cruising speed of a typical jet is 478 knots. We also learn that Clippy is 3.5 cm tall.

888.88 km/h are very, very lucky speed digits for Chinese people. Any aircraft manufacturer or airline would benefit from promoting this cruising speed...

It just needs Clippy™ waving itself around for complete & utter uselessness.

@FreeMan - It almost does - look at the Length setting - for small numbers it gives you the units in "paperclips".

@DarrelHoffman ugh... I was kidding!!!

@FreeMan: so was Microsoft. :-D

@FreeMan, Your suggestion is so popular that I had to include it!

I will do my best to refrain from down-voting simply for the presence of Clippy. :shudder: ;)


text:
Is there any way for the baggage carrier to add some fake luggage?
For the sake of a story that I am writing, is there any way that a conspiracy luggage carrier could add some luggage to given aircraft?
Specifically:

Is there still a man power used for the purpose of transporting luggage from the check-in to luggage trolley?
If yes, how do any of them know which luggage transportation belt they are supposed to deliver particular luggage to?
Are man present at all anywhere in the luggage sorting and transportation zones or is this today a fully automated process?

I am asking about a large international airport / Heathrow, if that matters.

The exact process varies quite a bit from airport to airport and even sometimes from terminal to terminal within an airport. But assuming a relatively modern terminal at a relatively busy airport, most of the process is heavily automated to reduce both labor and error.
Bags will be taken by conveyor to a security station, where a human watches them pass through various scanners. Then they get sorted by machines onto belts going to various parts of the terminal, and then belts going to each gate. This is why bags all get a big white tag with one or more barcodes on it, connected to a database that (hopefully) knows where each barcode should be sent.
Once the bag arrives at the gate, a human will transfer it (just a few feet) from the belt either onto a baggage cart (for smaller planes) or into a Unit Load Device (for larger planes), where it then waits to be loaded onto the plane. Notably, all bags on the same cart or in the same ULD will be waiting for the same plane.
At the other end, a human transfers the bag from the cart or ULD onto to the belt, and it goes back up into the system to be scanned and sent either to a different gate or off to baggage claim.
A single baggage handler can't really change much. They could intentionally misload a bag on the wrong aircraft (heck, they do it often enough by accident), but that is only easy if the "right" aircraft were at the same gate, or at least near enough to move the bag without another handler noticing the odd behavior, and there's no guarantee they'd be assigned to that gate at the right time since baggage crews move around a terminal pretty fluidly as planes come and go. And the "mistake" would be quickly detected when the plane was unloaded at its destination, unless of course a handler there was part of the same conspiracy. Many airline employees have been caught smuggling drugs this way, and presumably many more haven't been caught, so it's completely realistic.
More practical for a true conspiracy, though, would be planting, bribing or blackmailing someone who can simply change the database that says where a particular barcode should go. There are systems that check to ensure a passenger and their luggage aren't separated, so you'd need to attach the bag to someone who you'd be certain wouldn't miss the flight, and then detach it at the other end so you didn't get caught.
Overall, it is a lot simpler to trick "the system" into unwittingly doing your nefarious work for you rather than having a grand conspiracy of humans trying to outwit said system. But it may not make as good of a story.

Thank you. A perfect answer for both me, writing my story, and anyone else willing to learn more about how does the baggage transportation systems inside the airport works.

I would think that a loud "Carp, how'd this get here??!?!?" would reasonably cover manually moving a bag from one belt/cart to another and probably wouldn't raise too much suspicion. Especially in the early days at DIA.


text:
What are CP-like codes or designations after SAS or Scandinavian flight attendant names?
While browsing through Wikipedia I came across this photograph:

It's description says (emphasis mine):

SAS DC-9, interior of cabin, service on board. Air hostess Kirsten Andersen and Birthe Bymose both CPHCV serves lunch to Elisabeth Pinske CPHKA and Torben Wittrup CPHZC-S. 1960s

I was wondering what these codes are used for or that is their meaning.

CPH = Copenhagen / København?

My guess would be some sort of technical employment titles. The CPH part is definitely for Copenhagen. CV could be for Cabin ... something? KA could be Kunde Ansvarlig (responsible for customers) or something. ZC-S ... no idea. S could be for Sales maybe

https://www.flickr.com/photos/sas-museum/48768143153/

Seems like first three letters are indeed for denoting Copenhagen (as SAS had and have more than just one hub) and the remaining one seems to be Dutch designation for some sort of level, employment etc.  The real question (now) in here is -- whether such question has any value or should be closed? It seems not possible to answer it.

I emailed the SAS museum in Oslo to inquire about the acronyms in this photo description. It took awhile, but I finally got the following reply:

I'm a retired SAS-captain volunteering at our museum. These codes are
IATA communication codes. CPH is Copenhagen airport, CV is the name of
a SAS-sub group, the cabin attendant's group there. So if I were to
send a old fashioned telex to that group, I would include a NAME with
address CPHCV. Similarly, my former address would be my NAME, OSLOA.
These addresses were used within almost all airline companies, an
external message to me would be addressed NAME, OSLOASK, where SK
denotes SAS.
CPHKA is the address of the administrative SAS-department in charge of
all SAS stations within Denmark. CPHZC-S is unknown to me, but -S
indicates a further sub-division.
We used to have mail departments at all SAS stations, and this type of
addressing was extensively used on envelopes and in phone-directories.
These days, we still have mail departments, but these are greatly
reduced since e-mail has taken over communication.
You'll find more info on internet:
https://en.wikipedia.org/wiki/IATA_airport_code and:
https://www.iata.org/en/services/codes/
Best of luck with your studying.
Leif Leren, at the SAS-museum at OSL.


text:
Optimizing the airplane for best cruising altitude and velocity
In the conceptual phase of the design, I am optimizing various entities of an airplane for the best cruising altitude and speed to reduce the cost, which also includes MTOW and fuel weight. Consider a scenario where airplane is undertaking a mission which will burn the entire mission fuel estimated. Now, in-flight it is very much likely that airplane is not able to fly at the best cruising altitude and/or speed. This will result in increase of fuel burn which might not be enough. How to tackle this problem of estimating fuel load for off-design conditions while optimizing the airplane?
One way to address this issue is to rely on the reserve fuel estimated for reserve mission such as loiter time. But, it might happen that airplane has to rely on the reserve fuel and due to increase in fuel burn during cruise, reserve fuel is less for loiter time. I don't think this is a desirable condition. Another way is to optimize the airplane for worst scenario. Now, the question arises what is that worst scenario. Is it flying at the cruise ceiling (cruising altitude and cruise ceiling are different) or something else?
Let me know how to address this problem. Thanks in advance!

Very few airplanes have been used the way they were designed for.
From the trainer B.E.2 (which was abused for reconnaissance missions early in WW I) to the interceptor F-104 (which was used as a ground attack plane later), there are numerous examples. And for every mission that is still within the capabilities of one design you can think of numerous others that aren't. Therefore, dimensioning for the worst possible mission is an exercise in futility.
If an airplane carries too much fuel, it will be less economical. If it has too much strength built in, its structure will be too heavy. If the engines are more powerful than needed, it will have poorer fuel economy. In the end, every design can only fill its niche. But there are some principles which make some designs more useable:

Since regulations require a minimum number of flight attendants depending on the number of passengers, small passenger airplanes come with 19 or 50 seats but rarely with 23 or 55.
To be able to fly all domestic US routes, the more successful designs will have enough range to fly to Hawaii from airports at the West coast. If maximum range with full payload falls shorter than that, the design will be less versatile for US carriers.
Since air temperature is constant in the stratosphere, the most economical flight altitude is in the tropopause. Fly lower and you will not make full use of the cooler air above, fly higher and wing size and engine thrust will be less than optimal. Exceptions apply for business jets which like to fly above all other traffic.
Redundancy requires at least two engines, and maintenance cost demand to keep the number of engines as low as possible. Therefore, except for the really large airliners, the most successful ones all have just two engines.
In order to offer one of the fastest connections on a specific route, the reference cruise speed of long-range airliners is traditionally set at Mach 0.85. Fly faster and Mach effects will decrease fuel efficiency (demanding thinner airfoils and/or more wing sweep), fly slower and your design will drop from the first page of connections.

If you wonder how much a deviation from optimum airspeed will reduce range, I plotted that for a subsonic design (transsonic airliners will bump into steep drag increases with only a small increase in Mach, so this diagram does not apply to jets). I used a quadratic polar which gives a good approximation of reality.

Range reduction over airspeed. Blue line: L/D at the specified speed. Red line: Range relative to optimum. The optimum speed in this case is 84 m/s. I also assumed constant engine efficiency over speed, so range purely depends on L/D.

One also needs to factor IFR reserve regulations into any range calculation.

Are you designing a new airplane? If I understand the question correctly, For a normally aspirated engine, the engine is most efficient when the throttle is fully open at a particular altitude, where it produces approximately 65% -75% power. Therefore the airframe with respect to wing design, angle of incidence etc should be matched to give the best lift/drag ratio at that altitude. You will have to consult with the engine manufacturer to establish what that altitude is. Your choice of propeller is also going to have to be factored.
Then you calculate the fuel reserves accordingly.
Fuel reserves are calculated according to legal requirements of the country in which the airplane is registered and also in which it is operating. This varies according to the type of aircraft and the operation. For instance the minimum legal reserves are different for a passenger jet on a commercial flight, to a single engine airplane being operated on a VFR flight in the private category. Most countries laws are similar but there might be small variations, and they all have to satisfy ICAO standards if the country is an ICAO member state.

thanks for your answer! I am just curious to know how airplane designers estimate reserve fuel in the initial phase of the design. I have a rough approximation of how thrust and efficiency varies with altitude and speed. The plane is optimized for the best cruising condition, including reserve fuel which is not for off-design condition. This leads to least fuel, but there should be some extra amount of fuel to accommodate off-design cruise conditions, right? So, I wanted to how to do that.

How do the airplane designers estimate additional fuel needed, along with the fuel for the best condition? If the fuel provided is for only for the best condition, then there won't be any space for additional fuel

@Pavan Use the fuel reserves mandated by the prospective operator. For airliners that would be detour to the nearest alternative plus 30 min loiter. In the end, range is what is possible without reserve fuel, including expected winds aloft. Tank capacity is usually large enough so that payload can be traded for additional fuel.

@PeterKämpf, thanks for your answer! I am considering the reserve fuel mission. I was doubt whether that would be enough or not.


text:
Procedure for simulating nonlinear flight dynamics
I am trying to simulate F/A-18 Hornet on Simulink. So, here is how I have made the model:

Firstly, the initial flight parameters(Alpha, beta, V(magnitude), rho) are being calculated based on the initial values of the states.
Then, these parameters, along with initial values of the states, are used for calculating the aerodynamic forces and moments.
As the Aerodynamic forces and moments are in the Wind Axes, I transformed them from Wind to Body using alpha and beta.
Then, the gravity force, which is transformed from the NED Frame to Body frame, is added to the transformed aerodynamic force.
Similarly, the thrust force is added, assuming that thrust is only in the X-direction( Body Frame), and no moment is generated because of it.
Then, total force and moment are sent into the Body Axes 6-DOF Quaternion Block present in Simulink, which gives me the states for the next iteration and process continues.

Please point any mistakes you find out in the above-mentioned iteration process. Now, for checking if the model is correct or not, I planned to give initial conditions as a trim condition, and if the model is correct, the plane should fly in the stable level flight. The trim point is:
 V = 300.92 ft/s
 [phi, theta, psi] (in rads) = [0, 0.1745, 0] (pitch is 10 degrees)
 [p, q, r] = [0, 0, 0]
 alpha = 0.1745;
 beta = 0;

On giving these initial conditions, the plane flys at a stable level flight. So, is this process of verifying the model enough? If not, then please outline the process which I can use to check my model. Thanks for the help!
PS: Reference for the Aerodynamic Model of FA-18 Hornet:- A. Chakraborty, P. Seiler, and G. J. Balas, “Susceptibility of f/a-18 flight controllers to the falling-leaf mode: Linear analysis,” Journal of guidance, control, and dynamics, vol. 34, no. 1, pp. 57–72, 2011.

Oh man, that  question seems beyond the scope of what this particular on-line format is really optimized for, but hope you get some useful answers nonetheless.

Oops! Do you know any online platform where these types of questions can be asked?

Didn't mean to discourage you from asking here; I've seen a few questions that were somewhat similar.  Sorry can't help w/ suggestions of where else to ask.

Most aerodynamic database is in stability axis. Double check that yours is in wind axis. Not sure what you mean by "checking if your model is correct". Which flight regimes you want to check and to what fidelity? Most deviations have to do with your aerodynamic, landing gear, and propulsion models.

@JZYL I have mentioned the reference paper in my question. The paper gives an aerodynamic model to calculate CL, CD, CY, Cl, Cm, and Cn. I know that all these coefficients are defined in the wind axes "only" (correct me if I am wrong).  So does that mean that aerodynamic model is in Wind Axes?
I want to simulate an event of elevator jam. Can you please elaborate on the deviations you are talking about in the last line? What types of deviations and from where?

I am not a flight dynamics expert, but let me give an extremely general answer on verification of a non-linear dynamic system.  consider that you have the non-linear system $\dot x =f(x)$.
This system may have multiple equilibrium points, which are the $x$ values such that $f(x)=0$.  You have checked that one of the desired equilibrium points is actually an equilibrium point in your simulation. So that's good.  But does your system have more than one equilbrium point?  if so, you probably want to check all of them.
Next, you want to check the stability of the equilibrium points.  You've checked that $f(x)=0$ for one equilibrium point.  You need to check what happens for a small perturbation say $x_0= x+\epsilon$.  If the equilbrium point is supposed to be stable, then you need to check that after some time, the system has returned to equilibrium.  If the equilbrium point is supposed to be unstable, you need to check that the system has diverged from the equilbrium point.
But even that is not good enough.  You also need to check the rates of convergence or divergence.  i.e. near the equilbrium point, you can linearize the system and find the eigenvalues.  i.e. if you are at a small perturbation away from a stable equililbrium, you expect the system to approach equilibrium at a certain rate, say $x=x_0 e^{-\lambda t}$.  You need to verify that the simulation matches the expected rate.
So to summarize, you need to verify that your simulation has the correct number of equilibrium points, that the stability of each equilbrium point is correct, and that the eigenvalues near each equilbrium point are correct.

Thanks for explaining a more robust way of verifying my model!


text:
Dual Ignition Systems
Is it possible to back up a single magneto ignition system using an electronic system on the same spark plugs rather than having to change to two plg heads?

Possible, or legal?  Certified, or homebuilt?

Are you wanting to do this just from a redundancy standpoint?

What do you do if the single mag somehow fails a cylinder in a shorted manner? Or if a plug gets fouled and stops firing? That backup system on the same mag will no do you any good.  Stick with 2 plugs per cylinder.

It's a BGA registered TMG.  I'm just looking to back-up the magneto in case of failure but without the complications of replacing the cylinder heads  (which I agree would be the ideal solution)..

The spark plug is a fairly minor part as an ignition failure risk factor.  Far more likely on the primary side or at the coil.  So dual source to single plugs gets you probably 80% of where you would be with dual plug heads.  In the Corvair world, it's theoretically possible to modify the heads for dual plugs, but is considered not worth the trouble considering the other failure modes introduced by making more holes in the head.  If you rummage around on William Wynne's site for a bit you should be able to stumble onto sourcing for the ignition diodes. I'm pretty sure he sells them.

@ Dick Skinner It sounds like you want to add a second ignition to a Limbach VW powered motorglider?

For a homebuilt or ultralight, yes.  There are high voltage diodes that take a high tension lead from 2 separate sources with a single output, made for racing, that allow you to run a single plug from either of two sources, whatever they are.  They are popular on Corvair aircraft engine conversions.   You can run a mag and battery system together, or two battery systems, or whatever, with a one-or-the-other ignition switch.


text:
What are the advantages of centerline armament?
In the ME 109, Messerschmidt went through a lot of effort to squeeze a cannon into the fuselage firing between the piston banks of the DB engine, exiting the spinner. Why not just put the cannons in the wings? Is it because the trajectory of a center-line shot doesn’t need to converge (as wing-mounted guns do) and therefore has a longer range of sight?

Yes it's because of dispersion.  With wing mounted guns you normally "harmonize" them to converge at a point, say 300 yards or 600 yards to have a sweet spot where most of the projectiles are passing through a small area to maximize the fire concentration.
But at ranges closer and farther, the streams of projectiles are more dispersed and their effects are reduced.  Watch gun camera footage of strafing runs in something like a P-51, and you can see the bullet strikes widely dispersed and coming together into a tighter cluster as the range closes in.
Putting the armament in the nose eliminates the range issue and you have maximum concentration at any range.  The later versions of the 109 with the Mk108 30mm cannon were just about ideal for this because the cannon was very compact and had a really high rate of fire for its caliber (about 4-600 rpm), so a single cannon + two MGs was plenty of firepower and was probably the ideal configuration for the 109.
The ME262 jet had 4 Mk108s in the nose, providing close to the concentrated firepower of the A-10's 30mm 3000 rpm gattling gun.  A very short burst would bring down any fighter if it scored hits.

In addition, the reaction forces from firing the nose-mounted gun(s) are on the centerline of the plane, which means that firing the guns does not produce a yawing moment if for example one of the wing-mounted guns gets jammed or runs out of ammunition.

In the case of the P-38, the F-22 of its day, you also had no torque effects from the counter-rotating engines, making it jet like in the lack of torque effects when adjusting power.  It was probably the best gun platform of any piston fighter.


text:
Possible reason for diverting domestic flight to a distant country
I have just spotted something weird (at least to me) on Flightradar24. A Lufthansa domestic flight from Munich to Frankfurt flying toward... Warsaw in Poland:

My first, and quite obvious conclusion was a Flightradar24 error, as it happens from time to time. But then, I have opened flight LH121 recent details and to my extreme surprise it turned out, that the very same flight was diverted to Warsaw yesterday as well:

Is there any logical, business or technical explanation for this? Is there any logic at all in diverting local, domestic flight to a distant airport in not so close country? Because from mine -- real layman to airline operational rules -- this seems very, very wrong.
I mean, they have like about ten other airports in Germany, about six more in Czech Republic, not mentioning next six or seven in Poland itself, that this plane could land on, before reaching Warsaw.
Not mentioning for the fact, that looking at Flightradar24 track it seems like this plane would pointing Warsaw from the very beginning of takeoff. I mean -- Frankfurt is exactly opposite the way.
What am I missing?

So Flightradar24 had the same bug today as it had yesterday?  According to Google, it didn't divert at all.  According to Lufthansa, it arrived as scheduled also.

I'm voting to close this question as off-topic because it seems to be asking about a software bug in a 3rd party service.

@Simon Taking the fact, that Google takes its data from just the operational sources, while Flightradar24 uses actual ADS-B broadcast, intercepted and pushed in by many separate receivers, I'd more like into believing, that FR24, not Google, is correct and that this particular plane has actually flown over my house. I've heard (and spotted myself) many strange errors in FR24, but never heard about bug causing to show a plane, that is flying for 2+ hours, while it actually flown for less then hour in a bit different part of world. But.. I'm a laymen to this, so I may be very wrong.

Nope.  Google takes flight status from the airline databases.   For the absolute authority, check the Lufthansa arrivals.  According to the airline, it's aircraft landed in FRA, as scheduled.  I doubt that they are lying.

Frankfurt airport says that it is landed on schedule. I trust more the airport than a third party. http://i.imgur.com/UdqlomB.png

Also according to Lufthansa the flight was not diverted. You can go to this site and type in the flight number to get details.

OK, so it seems like a bug in FR24. Then it is really OT. Sorry. Though you should admit, that this one is really weird one. BTW, @Federico Wouldn't be easier to post a link instead of making a screenshot, cropping it and posting to imgur? :>

No problem.  The probable explanation (myth plausible) is that Flightradar24 has got the wrong call sign associated with the airframe number and that track is for a different flight altogether.

@Simon That was my first thought (expressed in question itself). However, finding out two exactly the same errors on exactly the same flight on two continuing days has just made my quite not so sure.

@trejder, since FR24 uses ADS-B, it is unlikely that it would be showing a flight track where there was no aircraft. But it might have easily assigned wrong flight plan to the aircraft. The ADS-B only broadcasts the aircraft unique identifier and the flight plan has to be taken from some other source and matched. And the matching could have gone wrong because of bug, error in data entry to that system or because the airline switched equipment and the update didn't propagate or didn't propagate yet.

The difference in equipment for the two days showing a "diversion" is also a good indicator that this was actually a different flight. FlightAware shows that the actual DLH121 still used A320-family aircraft on these days, as per the usual (and that it didn't divert.)


text:
How long is the FDR data kept for?
I was more than sure that each subsequent flights causes to overwrite earlier FDR and CVR recordings. How then should I understand the following part (emphasis mine) from this Wikipedia article:

The National Transportation Safety Board downloaded data from the aircraft's recorder and found it was a habit: 98% of the previous 175 takeoffs were made with incomplete flight-control checks. The National Business Aviation Association analyzed 143,756 flights in 2013-2015 by 379 business aircraft and only partial flight-control checks were done before 15.6% of the takeoffs and no checks at all on 2.03% of the flights.

Is it truly possible that single FDR unit kept 175 past flight's data? Is it possible that NBAA was able to retrieve approx. 379 earlier flights' data from 379 planes (143,756 / 379 ~= 379), spanning even two years into the past?
What am I missing here? How long data is kept and not overwritten in single FDR unit?

The amount of data that an individual FDR contains, while dependent on the particular unit, is fixed, and would correspond roughly to some amount of flight time.  How far into the past that extends, and how many flights it captures, depends on how often the aircraft flies and how long the flights are.  If aircraft 1 is operated for a single long flight every few days, and aircraft 2 makes 10 short flights daily, the memory in the FDR in the second aircraft would contain more flights, but going back fewer days on the calendar, than a comparable FDR in the first aircraft (before the data is getting over-written).
However...
At major airlines, and probably many smaller operators, the FDR is downloaded regularly by maintenance as part of a FOQA (Flight Operations Quality Assurance) program, and the data is stored so it can be analyzed for trends over time.  This way, if some worrisome event comes to light, it's possible to have a server go back through data across a fleet of perhaps hundreds of aircraft and years of operations to see how often something similar happened, or almost happened but wasn't noticed.
Thus, to take the example from the OP, if somebody noticed that flight control checks weren't being performed correctly, a FOQA analyst could write a query to determine if a particular flight control check was "good" or "bad", and then have the query run against the data from tens or even hundreds of thousands of flights.  If that found that most flight control checks were done correctly, they might go further to see if the bad flight control checks correlated with something... short turn arounds, or late night flights (i.e. tired crews getting sloppy), or something else.
Then, instead of a message to the crews "hey, make sure you do your flight control checks", the message could be more specific: "Even on quick turns, it's important to make sure you get full travel on the controls, because we've seen XXX occur even when the aircraft was only parked for a few minutes..."  That sort of a specific, focused message tends to affect behavior more than "Hey-- do your job right, darn it!" messages do.
Thus, data like what the NBAA referred to in the OP doesn't necessarily have to all be contained in the FDR; it may have been downloaded & stored over time instead.  Data for 175 flights from one aircraft seems like it could have been all in memory; if that were a couple hundred hours of recording, that doesn't seem like an unreasonable amount of memory to put in a FDR.  On the other hand, the analysis of data from over 100,000 flights over two years is almost certainly from data that had been downloaded regularly over that span of time.


text:
How can a pilot involved in serious accident remain unknown?
Somewhat similar to this question. A short part from Wikipedia article about Emirates Flight 521 (emphasis mine):

The captain was an unnamed 34-year-old UAE national who had been with Emirates since March 2001 and had logged 7,457 flight hours, including 5,123 hours on the Boeing 777.

In which situations or under which circumstances a name of pilot involved in serious crash may remain unknown or hidden?

"unnamed" is not the same as "unknown".

Aviation incident investigations are not there to appoint blame, so specific names are irrelevant to incident investigations. The only time, generally, individuals will be named in relation to an incident or accident is if the case somehow goes to criminal court, which would, generally, only happen if someone deliberately caused the accident.

@expeditedescent or they do something awesome, which is why we all know the name "Sully"

@expeditedescent Hm... a bit weird claim, I must say... I've read about hundred or two hundreds of crash reports (on Wikipedia only, true) and this one is the first where names were not revealed...

@trejder That's probably not indicent investigation reports you've been reading then. The media may well dig up the name of the people involved, but such information almost never comes from accident investigation boards.

In this case, it's most likely that the source for the story (the airline?) simply chose not to name the pilot and the editor of the Wikipedia article didn't seek another source that might have done so.
It's virtually certain that the pilot's name appears in the official reports of the incident.  The airline likely chose not to give the name for PR or legal reasons.

Note that NTSB reports generally do not name mishap pilots any longer (they used to at one point in time, but that practice was discontinued decades ago, probably sometime in the 80s or 90s),